We know that f1: R → R, given by f1(x) = x, and f2(x) = -x are surjective functions.
Proving f1 is surjective :
Let y be an element in the co-domain (R), such that f1(x) = y.
f1(x) = y
x = y, which is in R.
So, for every element in the co-domain, there exists some pre-image in the domain.1(x)=f1(y)x=y
So, f1is surjective .
Proving f2 is surjective :Let f2(x)=f2(y)−x=−yx=y
Let y be an element in the co domain (R) such that f2(x) = y.
f2(x) = y
x = y, which is in R.
So, for every element in the co-domain, there exists some pre-image in the domain.1(x)=f1(y)x=y
So, f2 is surjective .
Proving (f1 + f2) is not surjective :
Given:
(f1 + f2) (x) = f1 (x) + f2 (x)= x + (-x) =0
So, for every real number x, (f1 + f2) (x)=0
So, the image of every number in the domain is same as 0.
Range = {0}
Co-domain = R
So, both are not same.
So, f1 + f2 is not surjective.